Chapter 13.1, Problem 50E

a.

To determine

To calculate: The number of different ways to arrange three couples for a musical show.

Answer to Problem 50E

The number of different ways to arrange three couples for a musical showare $720$ .

Explanation of Solution

Given information:

There are 3 couples that have reserved seats for a musical show.

Formula used:

If there are n objects taken n at a time then permutation is defined as $P\left(n,n\right)=n!$ .

Calculation:

Consider the provided information that there are 3 couples that have reserved seats for a musical show.

In order to arrange these 3 couples, use permutation as selection matters.

Recall that if there are n objects taken n at a time then permutation is defined as $P\left(n,n\right)=n!$ .

Here n is 6.

Required number of ways are $P\left(6,6\right)$ .

  $\begin{array}{c}P\left(6,6\right)=6!\\ =720\end{array}$

Thus, the number of different ways to arrange three couples for a musical show are $720$ .

b.

To determine

To calculate: The number of different ways to arrange three couples for a musical show if two members of each couple are to be sit together.

Answer to Problem 50E

The number of different ways to arrange three couples for a musical show if two members of each couple are to be sit togetherare $48$ .

Explanation of Solution

Given information:

There are 3 couples that have reserved seats for a musical show.

Formula used:

If there are n objects taken n at a time then permutation is defined as $P\left(n,n\right)=n!$ .

Calculation:

Consider the provided information that there are 3 couples that have reserved seats for a musical show.

In order to arrange these 3 couplesif two members of each couple are to be sit together.

Now, three couples can be arranged in $3!$ number of ways.

And two members of each couple can be arranged in $2!\cdot 2!\cdot 2!$ number of ways as there are three couples in total.

Total number of ways are,

  $\begin{array}{c}3!\cdot 2!\cdot 2!\cdot 2!=6\left(2\right)\left(2\right)\left(2\right)\\ =48\end{array}$

Thus, the number of different ways to arrange three couples for a musical show if two members of each couple are to be sit together are $48$ .